Transformation of Coordinate Systems

minderbinder
Messages
5
Reaction score
0

Homework Statement



Find a one-to-one C1 mapping f from the first quadrant of the xy-plane to the first quadrant of the uv-plane such that the region where x^2 \leq y \leq 2x^2 and 1 \leq xy \leq 3 is mapped to a rectangle. Compute the Jacobian det Df and the inverse mapping f^{-1}.

The hint for the question states: Map all the regions where ax^2 \leq y \leq bx^2 and c \leq xy \leq d to rectangles.

Homework Equations



I'm a little confused on what they mean by map to a rectangle.

The Attempt at a Solution



I'm at a loss of where to begin...
 
Physics news on Phys.org
Thought about this some more, and I think the solution should be:

(u, v) = f(x, y) = (\frac{y}{x^2}, xy)

I checked some coordinates and it appears to work. However, I got this solution through trial and error. Can someone point out to me a way to find the solution in a systematic way?
 
let's start with the second inequality. xy goes from 1 to 3. this forms one dimension of a rectangle - along the xy-axis. but instead of using an xy-axis, you could use a u-axis or v-axis, if you let u or v equal to xy. Hint hint.

now for the first inequality, if y goes from x^2 and 2x^2, is there a way to manipulate this so that the value of *some algebraic expression* goes from one integer to another? kinda like xy above?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top